Answer:
a. 211.7
Explanation:
Iron Pyrite reacts with Oxygen to produce Iron (II) Oxide and Sulphur (IV) Oxide.
The equation is as follows:
4FeS₂₍s₎ + 11O₂₍g₎ → 2Fe₂O₃₍s₎ + 8SO₂₍g₎
From the equation, 4 moles of FeS₂ produce 8 moles of SO₂.
Therefore the reaction ratio is 4:8 or 1:2
198.20 grams of FeS₂ into moles is calculated as follows:
Moles= Mass/RMM
RMM of FeS₂ is 119.9750g/mol.
Number of moles = 198.20/119.9750g/mol
=1.652 moles of FeS₂
The reaction ratio of FeS₂ to SO₂ produced is 1:2
Thus SO₂ produced = 1.652 moles×2/1=3.304 moles
The mass of SO₂ produced =Moles ×RMM
=3.304 moles ×64.0638 g/mol
=211.667 grams
=211.7g
The values of x represents that number of moles of water molecules that is present per mole of the salt magnesium sulfate. To determine the value for this, we need to know how much is the water that is lost after heating the sample assuming that all of the water molecules are evaporated leaving only the unhydrated form of the salt. We calculate as follows:
Mass of hydrated salt = 3.484 g
Mass after heating = 1.701 g
Mass lost = 3.484 g - 1.701 g = 1.783 g
The mass lost is equal to the mass of water lost.
Moles water lost = 1.783 g ( 1 mol / 18.02 g ) = 0.0989 mol H2O
Moles of unhydrated salt = 1.701 g ( 1 mol / 120.37 g ) = 0.0141 mol MgSO4
moles water / moles MgSO4 = 0.0989 mol H2O / 0.0141 mol MgSO4 = 7
Therefore, the value of x is 7.
In order to answer this question, the units of volume must be consistent. In this problem, we decide the unit m3 to be uniform. Option A is equal to 12 m3, option b is equal to 1.2x10^8/100^3 or 120 m3. Option C is 2.0 x10^4/ 10^3 or 20 m3. Option D is 1.2x10^8/ 1000^3 or 0.12 m3. The greatest volume is option b. 120 m3.
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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